Question

If x lies in the interval [0,1], then the least value of x2 + x + 1 is
(a) 3

(b)$\frac{3}{4}$

(c) 1

(d) none of these

Answer 1

$$\begin{gathered} \left(\mathrm{c}\right)1 \\ \mathrm{Given}:f\left(x\right)=x^2+x+1 \\ \Rightarrow f\text{'}\left(x\right)=2x+1 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 2x+1=0 \\ \Rightarrow 2x=-1 \\ \Rightarrow x=\frac{-1}{2}\notin \left[0,1\right] \\ \mathrm{At}\mathrm{extreme}\mathrm{points}: \\ f\left(0\right)=0 \\ f\left(1\right)=1+1+1=3>0 \\ \mathrm{So},x=1\mathrm{is}\mathrm{a}\mathrm{local}\mathrm{minima}. \end{gathered}$$